author | Kevin Walker <kevin@canyon23.net> |
Tue, 09 Aug 2011 19:28:39 -0600 | |
changeset 861 | 84bb5ab4c85c |
parent 560 | b138ee4a5938 |
child 902 | bb48ee2ecf9e |
permissions | -rw-r--r-- |
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put in a stub appendix for MoAM, but I'm going to go do other things next
Scott Morrison <scott@tqft.net>
parents:
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%!TEX root = ../../blob1.tex |
833bd74143a4
put in a stub appendix for MoAM, but I'm going to go do other things next
Scott Morrison <scott@tqft.net>
parents:
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833bd74143a4
put in a stub appendix for MoAM, but I'm going to go do other things next
Scott Morrison <scott@tqft.net>
parents:
diff
changeset
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\section{The method of acyclic models} \label{sec:moam} |
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Let $F_*$ and $G_*$ be chain complexes. |
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Assume $F_k$ has a basis $\{x_{kj}\}$ |
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(that is, $F_*$ is free and we have specified a basis). |
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(In our applications, $\{x_{kj}\}$ will typically be singular $k$-simplices or |
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$k$-blob diagrams.) |
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For each basis element $x_{kj}$ assume we have specified a ``target" $D^{kj}_*\sub G_*$. |
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We say that a chain map $f:F_*\to G_*$ is {\it compatible} with the above data (basis and targets) |
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if $f(x_{kj})\in D^{kj}_*$ for all $k$ and $j$. |
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Let $\Compat(D^\bullet_*)$ denote the subcomplex of maps from $F_*$ to $G_*$ |
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such that the image of each higher homotopy applied to $x_{kj}$ lies in $D^{kj}_*$. |
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\begin{thm}[Acyclic models] \label{moam-thm} |
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Suppose |
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\begin{itemize} |
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\item $D^{k-1,l}_* \sub D^{kj}_*$ whenever $x_{k-1,l}$ occurs in $\bd x_{kj}$ |
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with non-zero coefficient; |
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\item $D^{0j}_0$ is non-empty for all $j$; and |
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\item $D^{kj}_*$ is $(k{-}1)$-acyclic (i.e.\ $H_{k-1}(D^{kj}_*) = 0$) for all $k,j$ . |
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\end{itemize} |
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Then $\Compat(D^\bullet_*)$ is non-empty. |
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If, in addition, |
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\begin{itemize} |
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\item $D^{kj}_*$ is $m$-acyclic for $k\le m \le k+i$ and for all $k,j$, |
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\end{itemize} |
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then $\Compat(D^\bullet_*)$ is $i$-connected. |
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\end{thm} |
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\begin{proof} |
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(Sketch) |
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This is a standard result; see, for example, \cite[Chapter 4]{MR0210112}. |
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We will build a chain map $f\in \Compat(D^\bullet_*)$ inductively. |
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Choose $f(x_{0j})\in D^{0j}_0$ for all $j$ |
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(possible since $D^{0j}_0$ is non-empty). |
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Choose $f(x_{1j})\in D^{1j}_1$ such that $\bd f(x_{1j}) = f(\bd x_{1j})$ |
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(possible since $D^{0l}_* \sub D^{1j}_*$ for each $x_{0l}$ in $\bd x_{1j}$ |
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and $D^{1j}_*$ is 0-acyclic). |
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Continue in this way, choosing $f(x_{kj})\in D^{kj}_k$ such that $\bd f(x_{kj}) = f(\bd x_{kj})$ |
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We have now constructed $f\in \Compat(D^\bullet_*)$, proving the first claim of the theorem. |
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parents:
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parents:
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Now suppose that $D^{kj}_*$ is $k$-acyclic for all $k$ and $j$. |
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Let $f$ and $f'$ be two chain maps (0-chains) in $\Compat(D^\bullet_*)$. |
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Using a technique similar to above we can construct a homotopy (1-chain) in $\Compat(D^\bullet_*)$ |
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between $f$ and $f'$. |
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Thus $\Compat(D^\bullet_*)$ is 0-connected. |
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Similarly, if $D^{kj}_*$ is $(k{+}i)$-acyclic then we can show that $\Compat(D^\bullet_*)$ is $i$-connected. |
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\end{proof} |
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